वर्गमूल सर्पिल में यदि \(\sqrt{120}\) कर्ण पर (1) इकाई लंब बनाई जाए, तो नया कर्ण कौन-सा होगा और वह किस अंतराल में आएगा?
If a (1) unit perpendicular is drawn on hypotenuse \(\sqrt{120}\) in a square root spiral, what will be the new hypotenuse and in which interval will it lie?
Explanation opens after your attempt
A. \(\sqrt{121}\), (11) पर\(\sqrt{121}\), at (11)
Concept
The new hypotenuse is \(\sqrt{120+1}=\sqrt{121}\), and \(\sqrt{121}=11\). When a perfect square appears, write its exact value.
Why this answer is correct
The correct answer is A. \(\sqrt{121}\), (11) पर / \(\sqrt{121}\), at (11). The new hypotenuse is \(\sqrt{120+1}=\sqrt{121}\), and \(\sqrt{121}=11\). When a perfect square appears, write its exact value.
Exam Tip
नया कर्ण \(\sqrt{120+1}=\sqrt{121}\) होगा और \(\sqrt{121}=11\) है। पूर्ण वर्ग दिखे तो उसका सटीक मान लिखें।
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